Problem 27 - Calculus Early Transcendentals

Question

What differentiable functions have an arc length on the interval \([a, b]\) given by the following integrals? Note that the answers are not unique. Give a family of functions that satisfy the conditions. a. \(\int_{a}^{b} \sqrt{1+16 x^{4}} d x \quad\) b. \(\int_{a}^{b} \sqrt{1+36 \cos ^{2}(2 x)} d x\)

Step-by-Step Solution

Answer

a) The family of functions satisfying the arc length expression \(\int_{a}^{b} \sqrt{1+16x^4}dx\) is given by \(y(x) = \frac{4}{3}x^3 + C\) for any constant \(C\). b) The family of functions satisfying the arc length expression \(\int_{a}^{b} \sqrt{1+36\cos^2(2x)}dx\) is given by \(y(x) = 3\sin(2x) + C\) for any constant \(C\).

Step 1: Determine the expression under the square root

Given arc length: $ \int_{a}^{b} \sqrt{1+16x^4}dx $ Expression under the square root is \(1+16x^4\).

Step 2: Identify the derivative

From the arc length formula, we have: $$ 1 + \left(\frac{d y}{d x}\right)^{2} = 1 + 16x^4 \\ \DRightarrow \left(\frac{d y}{d x}\right)^{2} = 16x^4 \\ \DRightarrow \frac{d y}{d x} = 4x^2 $$

Step 3: Integrate the derivative

Integrating the derivative with respect to \(x\), we obtain $$ y(x) = \int 4x^2 dx = \frac{4}{3}x^3 + C $$

Step 4: Write the family of functions

The family of functions satisfying the given condition is $$ y(x) = \frac{4}{3}x^3 + C $$for any constant \(C\). #Case b:#

Step 5: Determine the expression under the square root

Given arc length: $ \int_{a}^{b} \sqrt{1+36\cos^2(2x)}dx $ Expression under the square root is \(1+36\cos^2(2x)\).

Step 6: Identify the derivative

From the arc length formula, we have: $$ 1 + \left(\frac{d y}{d x}\right)^{2} = 1 + 36\cos^2(2x) \\ \DRightarrow \left(\frac{d y}{d x}\right)^{2} = 36\cos^2(2x) \\ \DRightarrow \frac{d y}{d x} = 6\cos(2x) $$

Step 7: Integrate the derivative

Integrating the derivative with respect to \(x\), we obtain $$ y(x) = \int 6\cos(2x) dx = 3\sin(2x) + C $$

Step 8: Write the family of functions

The family of functions satisfying the given condition is $$ y(x) = 3\sin(2x) + C $$for any constant \(C\).

Key Concepts

Arc Length

When we talk about the arc length of a function over an interval \([a, b]\), we refer to the "curved" distance along a curve. It's like measuring the length of a string that follows the path of the curve. In calculus, to find this distance, we use an integral based on the derivative of the function.

The general formula for the arc length from point \((a, f(a))\) to \((b, f(b))\) is given by the integral:

• \( L = \int_{a}^{b} \sqrt{1 + \left(\frac{d y}{dx}\right)^2} \: dx \)

Understanding arc length is essential because it incorporates both the path of the curve and the growth of the function along the interval.

Derivative

Derivatives help us understand how a function changes as its input changes. They're like the speedometers in your car, showing how fast the function's value is changing at any given point.

In our exercise, by analyzing the arc length formula, we isolated the expression under the square root to get \(1 + \left(\frac{dy}{dx}\right)^2\). To find possible functions with the given arc length, we equated the derivative squared to match the expression:

• \( \left(\frac{dy}{dx}\right)^2 = 16x^4 \), which gives \( \frac{dy}{dx} = 4x^2 \)
• \( \left(\frac{dy}{dx}\right)^2 = 36\cos^2(2x) \), leading to \( \frac{dy}{dx} = 6\cos(2x) \)

This approach allows us to align derivatives with the conditions provided and find functions whose derivatives match these specific forms.

Integration

Integration is the process of finding a function given its derivative. It's like working backward from knowing your speed at every point along a trip to figuring out where you started and ended.

For our arc length problem, having determined the derivatives, we integrate to find the original functions:

• From \( \frac{dy}{dx} = 4x^2 \) we integrate to get \( y(x) = \int 4x^2 \: dx = \frac{4}{3}x^3 + C \)
• From \( \frac{dy}{dx} = 6\cos(2x) \), we integrate to get \( y(x) = \int 6\cos(2x) \: dx = 3\sin(2x) + C \)

In both cases, \(C\) represents an integration constant, capturing the idea of a possible vertical shift in the graph of the function.

Family of Functions

A family of functions is a group of functions that share a common structure but can vary by certain parameters. In calculus, when we solve a differential equation or perform integration, we often encounter arbitrary constants like \(C\).

These constants lead to multiple solutions that form a family of functions. For instance:

• The family \( y(x) = \frac{4}{3}x^3 + C \) includes curves like \( y(x) = \frac{4}{3}x^3 + 1 \), \( y(x) = \frac{4}{3}x^3 - 2 \), etc., depending on \(C\).
• The family \( y(x) = 3\sin(2x) + C \) would include functions like \( y(x) = 3\sin(2x) \), \( y(x) = 3\sin(2x) + 5 \), etc.

This diverse set of solutions allows us to describe a wide range of graphs all derived from those specific conditions.